Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace rel_ops;
using ll = int64_t;
using Pii = pair<int, int>;
using ull = uint64_t;
using Vi = vector<int>;
void run();
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  run();
  return 0;
}
using Point = pair<ll, ll>;
ll sY, a, b;
int n;
vector<Point> segments;
vector<ll> prefs;
Point cur;
double proj(Point point, ll first) {
  double alfa = double(point.first - first) / double(point.second);
  return alfa * double(sY) + double(first);
}
bool projCmp(Point l, ll val) {
  return (cur.first - l.second) * sY + l.second * cur.second < val * cur.second;
}
void handle() {
  cin >> cur.first >> cur.second;
  int begin =
      int(lower_bound((segments).begin(), (segments).end(), a, projCmp) -
          segments.begin());
  int end = int(lower_bound((segments).begin(), (segments).end(), b, projCmp) -
                segments.begin());
  double ret = double(prefs[end] - prefs[begin]) * double(cur.second - sY) /
               double(cur.second);
  if (begin < n) {
    ret -= max(double(a) - proj(cur, segments[begin].first), 0.0);
  }
  if (end < n) {
    ret += max(double(b) - proj(cur, segments[end].first), 0.0);
  }
  cout << ret << '\n';
}
void run() {
  cin >> sY >> a >> b >> n;
  segments.resize(n);
  prefs.push_back(0);
  for (auto& s : (segments)) {
    cin >> s.first >> s.second;
    prefs.push_back(prefs.back() + s.second - s.first);
  }
  int q;
  cin >> q;
  while (q--) handle();
}